[英]How to sort array of integer first by value and second by number of repetition using swift in time complexity < O(n^2) and space complexity O(n)
这是我尝试过的解决方案,但是顺序为O(n ^ 2),所以没有通过测试结果
func sortArrayByValueAndByFrequency(nums : [Int]) {
var countDict = [Int : Int]()
var count = Int()
var values = Int()
var output = [Int]()
for index in 0 ..< nums.count {
for index2 in 0 ..< nums.count{
if nums[index2] == nums[index] {
values = nums[index2]
count += 1
}
}
countDict[values] = count
count = 0
}
let sortedByKey = countDict.sorted { ($0.key < $1.key)}
let sortedByValue = sortedByKey.sorted { ($0.value < $1.value)}
for (k,v) in sortedByValue {
for _ in 1 ... v {
output.append(k)
}
}
output.forEach { (orderedNumber) in
print(orderedNumber)
}
}
输入/输出示例:
Example array = [1,1,2,3,4,5,5,6,7,7,7,8,9,9,9,20,25,21,20]
Expected output = [2,3,4,6,8,21,25,1,1,5,5,20,20,7,7,7,9,9,9]
example 2 = [1,2,3,4,4,3,3]
output = [1,2,4,4,3,3,3]
这个问题是在HackerRank上问我的
首先确定每个值的出现次数(O(n)),然后对这些值进行排序,并以出现次数为第一排序标准,并将值本身作为第二排序标准(O(n log(n)) )。 通过元组比较方便地进行排序 (比较Swift-使用多个条件对对象数组进行排序 ):
let array = [1,1,2,3,4,5,5,6,7,7,7,8,9,9,9,20,25,21,20]
let countDict = array.reduce(into: [Int:Int]()) {
$0[$1, default: 0] += 1
}
let sorted = array.sorted(by: {
(countDict[$0]!, $0) < (countDict[$1]!, $1)
})
print(sorted)
// [2, 3, 4, 6, 8, 21, 25, 1, 1, 5, 5, 20, 20, 7, 7, 7, 9, 9, 9]
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